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\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2016\cdot2017}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(=1-\frac{1}{2017}=\frac{2016}{2017}\)
Ta có : \(x=\frac{2015.2016+2}{2015.2016}=\frac{2015.2016}{2015.2016}+\frac{2}{2015.2016}=1+\frac{1}{1008.2015}\)
\(y=\frac{2016.2017+2}{2016.2017}=\frac{2016.2017}{2016.2017}+\frac{2}{2016.2017}=1+\frac{1}{1008.2017}\)
Vì \(\frac{1}{1008.2015}>\frac{1}{1008.2017}\)
=> \(1+\frac{1}{1008.2015}>1+\frac{1}{1008.2017}\)
=> \(\frac{2015.2016+2}{2015.2016}>\frac{2016.2017+2}{2016.2017}\)
=> \(x>y\)
Ta có:
x = \(\frac{2015.2016+2}{2015.2016}=\frac{2015.2016}{2015.2016}+\frac{2}{2015.2016}=1+\frac{2}{2015.2016}=1+\frac{1}{2015.1008}\)
y = \(\frac{2016.2017+2}{2016.2017}=\frac{2016.2017}{2016.2017}+\frac{2}{2016.2017}=1+\frac{2}{2016.2017}=1+\frac{1}{1008.2017}\)
Do \(\frac{1}{2015.1008}>\frac{1}{1008.2017}\) => \(1+\frac{1}{2015.1008}>1+\frac{1}{1008.2017}\)
=> x > y
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2016.2017}\)
\(A=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+......+\left(\frac{1}{2016}-\frac{1}{2017}\right)\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2016}-\frac{1}{2017}\)
\(A=\frac{1}{1}-\frac{1}{2017}\)
\(A=\frac{2016}{2017}\)
A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2016.2017}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{2016}-\frac{1}{2017}\)
\(\Rightarrow A=1-\frac{1}{2017}\)
\(\Rightarrow A=\frac{2016}{2017}\)
A=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2016}-\frac{1}{2017}\)
A=\(\frac{1}{1}-\frac{1}{2017}\)
A=\(\frac{2016}{2017}\)
a.\(\frac{2015.2016-1}{2015.2016}=1-\frac{1}{2015.2016}\)
\(\frac{2016.2017-1}{2016.2017}=1-\frac{1}{2016.2017}\)
vì \(\frac{1}{2015.2016}>\frac{1}{2016.2017}\)
=>\(-\frac{1}{2015.2016}< -\frac{1}{2016.2017}\)
=>\(1-\frac{1}{2015.2016}< 1-\frac{1}{2016.2017}\)